有關Oracle邏輯讀和物理讀，你了解嗎？
1.物理讀(physical read)
物理讀即是把數據從磁盤讀入到buffer catch的過程。 通常情況下是，如果需要數據的時候發現不存在于buffer catch當中，即oracle就會執行物理讀。
當數據塊***次讀取到,就會緩存到buffer cache 中,而第二次讀取和修改該數據塊時就在內存buffer cache 了 以下是例子:
1.1 ***次讀取:
C:\Documents and Settings\Paul Yi>sqlplus "/as sysdba"
SQL*Plus: Release 9.2.0.4.0 - Production on Thu Feb 28 09:32:04 2008
Copyright (c) 1982, 2002, Oracle Corporation. All rights reserved.
Connected to:
Oracle9i Enterprise Edition Release 9.2.0.4.0 - Production
With the Partitioning, OLAP and Oracle Data Mining options
JServer Release 9.2.0.4.0 - Production
SQL> set autotrace traceonly
SQL> select * from test;
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT ptimizer=CHOOSE (Cost=2 Card=4 Bytes=8)
1 0 TABLE ACCESS (FULL) OF 'TEST' (Cost=2 Card=4 Bytes=8)
Statistics
----------------------------------------------------------
175 recursive calls
0 db block gets
24 consistent gets
9 physical reads --9個物理讀
0 redo size
373 bytes sent via SQL*Net to client
503 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
2 sorts (memory)
0 sorts (disk)
1 rows processed
1.2 第二次讀取
SQL> select * from test;
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT ptimizer=CHOOSE (Cost=2 Card=4 Bytes=8)
1 0 TABLE ACCESS (FULL) OF 'TEST' (Cost=2 Card=4 Bytes=8)
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
7 consistent gets
0 physical reads --沒有發生物理讀了,直接從buffer cache 中讀取了
0 redo size
373 bytes sent via SQL*Net to client
503 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1 rows processed
1.3 數據塊被重新讀入buffer cache ,這種發生在
如果有新的數據需要被讀入Buffer Cache中，而Buffer Cache又沒有足夠的空閑空間，Oracle就根據LRU算法將LRU鏈表中LRU端的數據置換出去。當這些數據被再次訪問到時，需要重新從磁盤讀入。
SQL> alter session set events 'immediate trace name flush_cache'; --清空數據緩沖區
Session altered.
SQL> select * from test;
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT ptimizer=CHOOSE (Cost=2 Card=4 Bytes=8)
1 0 TABLE ACCESS (FULL) OF 'TEST' (Cost=2 Card=4 Bytes=8)
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
7 consistent gets
6 physical reads --又重新發生了物理讀
0 redo size
373 bytes sent via SQL*Net to client
503 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1 rows processed
2.邏輯讀(buffer read)
邏輯讀指的就是從(或者視圖從)Buffer Cache中讀取數據塊。按照訪問數據塊的模式不同，可以分為即時讀(Current Read)和一致性讀(Consistent Read)。注意：邏輯IO只有邏輯讀，沒有邏輯寫。
即時讀
即時讀即讀取數據塊當前的***數據。任何時候在Buffer Cache中都只有一份當前數據塊。即時讀通常發生在對數據進行修改、刪除操作時。這時，進程會給數據加上行級鎖，并且標識數據為“臟”數據。
SQL> select * from test for update;
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT ptimizer=CHOOSE (Cost=2 Card=4 Bytes=8)
1 0 FOR UPDATE
2 1 TABLE ACCESS (FULL) OF 'TEST' (Cost=2 Card=4 Bytes=8)
Statistics
----------------------------------------------------------
0 recursive calls
1 db block gets
14 consistent gets
0 physical reads
252 redo size
386 bytes sent via SQL*Net to client
503 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1 rows processed
SQL>
一致性讀
Oracle是一個多用戶系統。當一個會話開始讀取數據還未結束讀取之前，可能會有其他會話修改它將要讀取的數據。如果會話讀取到修改后的數據，就會造成數據的不一致。一致性讀就是為了保證數據的一致性。在Buffer Cache中的數據塊上都會有***一次修改數據塊時的SCN。如果一個事務需要修改數據塊中數據，會先在回滾段中保存一份修改前數據和SCN的數據塊，然后再更新Buffer Cache中的數據塊的數據及其SCN，并標識其為“臟”數據。當其他進程讀取數據塊時，會先比較數據塊上的SCN和自己的SCN。如果數據塊上的SCN小于等于進程本身的SCN，則直接讀取數據塊上的數據;如果數據塊上的SCN大于進程本身的SCN，則會從回滾段中找出修改前的數據塊讀取數據。通常，普通查詢都是一致性讀。
下面這個例子幫助大家理解一下一致性讀：
會話1中：
SQL> select * from test;
ID
----------
1000
SQL> update test set id=2000;
1 row updated.
會話2中:
SQL> set autotrace on
SQL> select * from test;
ID
----------
1000
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT ptimizer=CHOOSE (Cost=2 Card=4 Bytes=8)
1 0 TABLE ACCESS (FULL) OF 'TEST' (Cost=2 Card=4 Bytes=8)
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
9 consistent gets 沒有事物做update時 是 7 consistent gets 說明多了2個 consistent gets 這2個是要從回滾段中獲取的
0 physical reads
52 redo size
373 bytes sent via SQL*Net to client
503 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1 rows processed
SQL>
